The mean of $5$ observation is $5$ and their variance is $12.4$. If three of the observations are $1, 2$ and $6$; then the mean deviation from the mean of the data is:

Let the missing observations be x and y. Since the mean is 5, the sum is 25, so 1 + 2 + 6 + x + y = 25, which gives x + y = 16. The variance is 12.4, so the sum of squared deviations is 5 times 12.4 = 62. This gives the equation (1-5)^2 + (2-5)^2 + (6-5)^2 + (x-5)^2 + (y-5)^2 = 62, which simplifies to 26 + (x-5)^2 + (y-5)^2 = 62, yielding (x-5)^2 + (y-5)^2 = 36. Let u = x-5 and v = y-5. Then u+v = (x+y)-10 = 6, and u^2 + v^2 = 36. This implies 2uv = 0, so either u=0 or v=0. Thus, the five observations are 1, 2, 6, 5, and 11. The mean deviation about the mean is (|1-5| + |2-5| + |6-5| + |5-5| + |11-5|) / 5 = (4 + 3 + 1 + 0 + 6) / 5 = 14/5. The result is 2.8.

• Thanks
• Wrong explanation